4
$\begingroup$

I have two lists, for example

s={{1,3,2},{1,3,2},{2,3,1},{3,1},{2,3,1}};
c={{3},{1,2},{3}};

First element of c means that number 1 from s[[3]] must be deleted. Number is a position of this element in c.

Second element of c: delete number 2 from s[[{1,2}]].

Third: delete number 3 from s[[3]]

The result must be

(* {{1,3},{1,3},{2},{3,1},{2,3,1}} *)

Another example

s={{2,1,3},{2,3,1},{1,2,3},{2,3}};
c={{1},{2,4},{3}};
(* {{2,3},{3,1},{1,2},{3}} *)

One more question. Also two lists:

s={{{1,3},{1,2},{3,2},{1},{3},{2}},
{{1,3},{1,2},{3,2},{1},{3},{2}},
{{1},{2},{3}},
{{3},{2},{1}}};
c={{3},{1,2},{}};

So, from s[[3]] (* {{1},{2},{3}} *) it's necessary to delete ALL elements which contain number 1: {1} (* {{2},{3}} *)

From s[[{1,2}}]] - all elements which contain number 2

(* {{{1,3},{1},{3}},
{{1,3},{1},{3}},
{{2},{3}},
{{3},{2},{1}}} *)

How can I do it?

Improve this question
$\endgroup$
0
8
$\begingroup$

You may use MapIndexed, ReplaceAll, and Nothing.

Delete Numbers

s1 = {{1, 3, 2}, {1, 3, 2}, {2, 3, 1}, {3, 1}, {2, 3, 1}};
c1 = {{3}, {1, 2}, {3}};

MapIndexed[(s1[[#1]] = (s1[[#1]] /. {First@#2 -> Nothing})) &, c1];
s1

(* {{1, 3}, {1, 3}, {2}, {3, 1}, {2, 3, 1}} *)

Delete Sublists

s2 = {
   {{1, 3}, {1, 2}, {3, 2}, {1}, {3}, {2}},
   {{1, 3}, {1, 2}, {3, 2}, {1}, {3}, {2}},
   {{1}, {2}, {3}},
   {{3}, {2}, {1}}
   };
c2 = {{3}, {1, 2}, {}};

MapIndexed[(s2[[#1]] = (s2[[#1]] /. {{___, First@#2, ___} -> Nothing})) &, c2];
s2

(*
{{1,3},{1},{3}}
{{1,3},{1},{3}}
{{2},{3}}
{{3},{2},{1}}
*)

Hope this helps.

Improve this answer
$\endgroup$
3
  • $\begingroup$ if possible can you tell more about the aim of author who asked the question I have so much tried to understand your solution but I think I cannot understand the main question! $\endgroup$
    – Unbelievable
    Jul 25 '16 at 13:29
  • $\begingroup$ @Irreversible OP has a list of positions c and wants to perform some action on the list s that is a function of each position and the position's index in list of positions. MapIndexed is mapping its function over c. MappedIndexed's function takes two arguments. The first is the item from the list it is mapping over (#1) and the second is the index of that item in the list (#2). #1 is obtained from the list. While #2 is generated by the system for you; you don't need to provide it. Try MapIndexed[{#1,#2}&, Range[6,10]]. $\endgroup$
    – Edmund
    Jul 25 '16 at 21:02
  • $\begingroup$ Thank you so much. OP has used of phrases such as : Number 1 Number 2!!! they are not defined in list-manipulation category. I wish a professional OP edits his/her question. $\endgroup$
    – Unbelievable
    Jul 26 '16 at 5:32
4
$\begingroup$

Do and DeleteCases can do what you want

The first one:

delete[s_, c_] := Module[{s0 = s},
  Do[s0[[c[[i]]]] = DeleteCases[s0[[c[[i]]]], i, 2], {i, Length@c}];
  s0
  ]

The second one

delete2[s_, c_] := Module[{s0 = s},
  Do[s0[[c[[i]]]] = DeleteCases[s0[[c[[i]]]], {___, i, ___}, 2], {i, 
    Length@c}];
  s0
  ]
Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.